Lesson 9
Using Tables for Conditional Probability
9.1: Math Talk: Fractions in Fractions (5 minutes)
Warm-up
The purpose of this Math Talk is to elicit strategies and understandings students have for dividing fractions. These understandings help students develop fluency and will be helpful later in this lesson when students will need to be able to divide one probability (often in the form of a fraction) by another probability.
In this activity, students have an opportunity to notice and make use of structure (MP7) when they recognize fraction bars as part of a fraction as well as representing division.
Launch
Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.
Supports accessibility for: Memory; Organization
Student Facing
Evaluate each expression mentally.
\(\frac{7}{11} \div \frac{8}{11}\)
\(\frac{\left( \frac{1}{3} \right)}{\left( \frac{2}{3} \right)}\)
\(\frac{\left( \frac{1}{4} \right)}{\left( \frac{1}{2} \right)}\)
\(\frac{\left( \frac{3}{8} \right) \left( \frac{8}{19} \right)}{\left( \frac{5}{19} \right)}\)
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
- “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
- “Did anyone have the same strategy but would explain it differently?”
- “Did anyone solve the problem in a different way?”
- “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
- “Do you agree or disagree? Why?”
Design Principle(s): Optimize output (for explanation)
9.2: A Possible Cure (15 minutes)
Activity
The mathematical purpose of this activity is to use a two-way table as a sample space to decide if events are independent and to estimate conditional probabilities.
Making spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Give students 10 minutes of quiet time to work the problems and then pause for a brief whole-class discussion.
Supports accessibility for: Organization; Conceptual processing; Attention
Student Facing
A pharmaceutical company is testing a new medicine for a disease using 115 test subjects. Some of the test subjects are given the new medicine and others are given a placebo. The results of their tests are summarized in the table.
| no more symptoms | symptoms persist | total | |
|---|---|---|---|
| given medicine | 31 | 26 | 57 |
| given placebo | 16 | 42 | 58 |
| total | 47 | 68 | 115 |
- Divide the value in each cell by the total number of test subjects to find each probability to two decimal places. Some of the values have been completed for you.
no more symptoms symptoms persist total given medicine 0.27 0.50 given placebo total 1 If one of these test subjects is selected at random, find each probability:
- \(P(\text{symptoms persist})\)
- \(P(\text{given medicine and symptoms persist})\)
- \(P(\text{given placebo or symptoms persist})\)
-
From the original table, divide each cell by the total for the row to find the probabilities with row conditions. Some of the values have been completed for you.
no more symptoms symptoms persist total given medicine 0.54 given placebo 1 - \(P(\text{symptoms persist | given medicine})\)
- \(P(\text{no more symptoms | given placebo})\)
- Jada didn’t read the instructions for the previous problem well and used the table she created on the first problem to divide each cell by the probability total for each row. For example, in the top left cell she calculated \(0.27 \div 0.5\). Complete the table using Jada’s method.
no more symptoms symptoms persist total given medicine given placebo What do you notice about this table?
-
From the original table, divide each cell by the total for the column to find the probabilities with column conditions. Some of the values have been completed for you.
no more symptoms symptoms persist given medicine 0.66 given placebo total 1 - \(P(\text{given medicine | symptoms persist})\)
- \(P(\text{given placebo | no more symptoms})\)
-
Are the events “symptoms persist” and “given medicine” independent events? Explain or show your reasoning.
-
Based on your work, does being given this medicine have an impact on whether symptoms persist or not?
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
Consider the data collected from all students in grades 11 and 12 and their intentions of going to prom.
-
65% of the students who are going to prom are from grade 12. Complete the table.
going to prom not going to prom total grade 11 127 grade 12 116 total 124 119 243 -
Based on your results, what is the probability that a student from this group, selected at random, is a student in grade 11 that did not go to prom?
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students might struggle with understanding the notation for conditional probability. Tell these students that \(P(\text{A | B})\) means the probability of A happening under the condition that B happens.
Activity Synthesis
The purpose of this discussion is for students to interpret probabilities in a two-way table.
Ask students to interpret each of the probabilities they found in questions 1, 2, and 4. For example, “What does the probability in question 1a mean in the context of the problem?”
Here are sample interpretations of the probabilities for questions 1, 2, and 4.
-
- The probability that a test subject chosen at random from all of the subjects still had symptoms after the test.
- The probability that a test subject chosen at random from all of the subjects was among those given medicine and still had symptoms after the test.
- The probability that a test subject chosen at random from all of the subjects still had symptoms after the test or was given a placebo. 1
-
- The probability that a test subject chosen at random from those who were given medicine had symptoms after the test.
- The probability that a test subject chosen at random from those who were given placebo had no more symptoms after the test
-
- The probability that a test subject chosen at random from those who had symptoms after the test was given medicine.
- The probability that a test subject chosen at random from those who had no more symptoms after the test was given placebo.
Display what Jada did when creating the top left cell in table for all to see:
\(\frac{P(\text{no more symptoms and given medicine})}{P(\text{given medicine})}\)
Here are some questions for discussion.
- “What do you notice about what Jada did and \(P(\text{no more symptoms | given medicine})\)?” (I noticed that they are equal.)
- “What is an equation that represents this situation using A and B for the events?” (\(P(\text{A | B}) = \frac{P(\text{A and B})}{P(\text{B})}\))
- "What is the relationship between this equation and the multiplication rule?" (From the multiplication rule, we know that \(P(\text{A and B}) = P(\text{A | B}) \boldcdot P(\text{B})\). If we divide both sides by \(P(\text{B})\), we get the equation that Jada used.)
- “How did you determine whether the events were dependent or independent?” (I determined that they were dependent events because \(P(\text{symptoms persist}) = 0.6\) and \(P(\text{symptoms persist | given medicine}) = 0.46\) were not equal. If the events were independent, these probabilities would be equal.)
Design Principle(s): Support sense-making
9.3: The Blood Bank (10 minutes)
Activity
The mathematical purpose of this activity is for students to use a two-way table to estimate conditional probabilities. Monitor for students discussing the difference between using “and” and “or.”
Launch
Arrange students in groups of 2. Give students quiet time to work on the questions, have partners compare answers, and then have a whole-class discussion.
Supports accessibility for: Language; Conceptual processing
Student Facing
A blood bank in a region has some information about the blood types of people in its community. Blood types are grouped into type O, A, B, and AB. Each blood type either has the Rh factor (Rh+) or not (Rh-). If a person is randomly selected from the community, the probability of their having each blood type and Rh factor combination is shown in the table.
| O | A | B | AB | |
|---|---|---|---|---|
| Rh+ | 0.374 | 0.357 | 0.085 | 0.034 |
| Rh- | 0.066 | 0.063 | 0.015 | 0.006 |
- What does the 0.085 in the table represent?
- Use the table or create additional tables to find the probabilities, then describe the meaning of the event.
- \(P(\text{O})\)
- \(P(\text{Rh+})\)
- \(P(\text{O and Rh+})\)
- \(P(\text{O or Rh+})\)
- \(P(\text{O | Rh+})\)
- \(P(\text{Rh+ | O})\)
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students might struggle with understanding the notation for conditional probability. Tell these students that \(P(\text{A | B})\) means the probability of A happening under the condition that B happens.
Activity Synthesis
The goal of this discussion is for students to understand how to estimate conditional probabilities using a two-way table and to informally assess student understanding of notation used with conditional probability.
Here are some questions for discussion.
- “What is the probability that a person selected is Rh- under the condition that they are blood type B?” (\(\frac{0.015}{0.1}=0.15\))
- “How would you write ‘the probability that a person selected is Rh- under the condition that they are blood type B’ using probability notation?” (\(P(\text{Rh- | B})\))
- “What is the probability that a person selected has blood type AB under the condition that they are Rh-?” (\(\frac{0.006}{0.15}=0.04\))
- “How would you write ‘the probability that a person selected has blood type AB under the condition that they are Rh-’ using probability notation?” (\(P(\text{AB | Rh-})\))
Lesson Synthesis
Lesson Synthesis
Display the table from the cool down and its description for all to see.
A large aquarium has 48 guppy fish in it. The table summarizes the fish based on color and sex.
| orange | blue | |
|---|---|---|
| male | 5 | 13 |
| female | 12 | 18 |
Here are some questions for discussion.
- “What is the probability that a fish picked at random is orange?” (\(\frac{17}{48}\))
- “What is the probability that a fish picked at random is orange under the condition that the fish is female? (\(\frac{12}{30}\))
- “What is the probability that a fish picked at random is orange and male?” (\(\frac{5}{48}\))
- “What is the probability that a fish picked at random is orange or male?” (\(\frac{30}{48}\))
- “What is the probability that a female fish picked at random is blue?” (\(\frac{18}{30}\))
- “How would you write ‘what is the probability that a female fish picked at random is blue?’ using probability notation?” (\(P(\text{blue | female}\))
- “Are the events of being an orange fish and being a female fish independent events?” (They are not independent events. Sample reasoning: \(P(\text{being an orange fish}) = \frac{17}{48}\) and \(P(\text{being an orange fish | being a female fish}) = \frac{12}{30}\). If the events were independent, these probabilities would be equal.)
9.4: Cool-down - Guppies (5 minutes)
Cool-Down
For access, consult one of our IM Certified Partners.
Student Lesson Summary
Student Facing
Organizing data in tables is a useful way to see information and compute probabilities. Consider the data collected from all students in grades 11 and 12 and their intentions of going to prom.
| going to prom | not going to prom | total | |
|---|---|---|---|
| grade 11 | 43 | 84 | 127 |
| grade 12 | 81 | 35 | 116 |
| total | 124 | 119 | 243 |
A student is randomly selected from this group of students. We can find a number of probabilities based on the table.
\(P(\text{grade 12 and going to prom}) = \frac{81}{243}\) since there are 81 students who are both in 12th grade and going to prom out of all 243 students in the group who could be selected.
\(P(\text{grade 12}) = \frac{116}{243}\) since there are 116 grade 12 students out of the entire group.
\(P(\text{going to prom}) = \frac{124}{243}\) since there are 124 students going to prom out of the entire group.
\(P(\text{grade 12 or going to prom}) = \frac{159}{243}\) since there are 159 students in grade 12 or going to prom (43 from grade 11 going to prom, 81 from grade 12 going to prom, and 35 from grade 12 not going to prom).
\(P(\text{going to prom | grade 12}) = \frac{81}{116}\) represents the probability that the chosen student is going to prom under the condition that they are in grade 12. The group we are considering is different for this probability since the condition is that they are in grade 12. Imagine all grade 12 students are in a room and we are only selecting from this group. Therefore 81 students are going to prom from this group out of 116 students in the group.
\(P(\text{grade 11 | going to prom}) = \frac{43}{124}\) represents the probability that the chosen student is in grade 11 under the condition that they are going to prom. In this case, we imagine that everyone going to prom is gathered in a room and we are selecting one student from this group. The probability that this student is in grade 11 when just considering this group is \(\frac{43}{124}\).
The last two probabilities can also be found using the multiplication rule.
\(P(\text{A and B}) = P(\text{A | B}) \boldcdot P(\text{B})\) can be rewritten \(P(\text{A | B}) = \frac{P(\text{A and B})}{P(\text{B})}\).
For example, substituting values into \(P(\text{going to prom | grade 12}) = \frac{P(\text{going to prom and grade 12})}{P(\text{grade 12})}\) gives \(\frac{81}{116} = \frac{\left(\frac{81}{243}\right)}{\left(\frac{116}{243}\right)}\) which is a true equation.